Question

Evaluate : 2sin 2 38 .sec 2 52- sin 2 45+ sec 2 54-cot 2 36 by cosec 2 57 -tan 2 33

Answer 1

(2sin²38°.sec²52°-sin²45°+sec²54°-cot²36°)/(cosec²57°-tan²33°)
={2sin²38°.sec²(90°-38°)-(1/√2)²+sec²54°-cot²(90°-54°)}/{cosec²57°-tan²(90°-57°)}
={2sin²38°cosec²38°-(1/2)+sec²54°-tan²54°}/(cosec²57°-cot²57°)
={2sin²38°(1/sin²38°)-(1/2)+1}/(1) 
[∵, sec²Ф-tan²Ф=1 and cosec²Ф-cot²Ф=1]
=2+1-(1/2)
=3-1/2 
=5/2 Ans.

Answer 2

Sin 38° = Cos (90° -38°) = Cos 52°
Sin 45 = 1/√2
Cot 36° = Tan (90° - 36°) = Tan 54°
Cosec 57 = Sec (90° - 57°) = sec 33°

LHS = 
$\frac{2 sin^2 38^o*sec^252^o-sin^2 45^0+sec^2 54^0-cot^2 36°}{cosec^2 57^o-tan^2 33^o}\\\\=\frac{2 cos^2 52^o*sec^2 52^o-1/2+sec^2 54^o-tan^2 54^o}{sec^2 33^o-tan^2 33^o}\\\\=\frac{2- 1/2+1}{1}=5/2$